Download Chemical Quantities

CHAPTER 9
Chemical Quantities
1.
The coefficients of the balanced chemical equation for a reaction give the relative numbers of
molecules of reactants and products that are involved in the reaction.
2.
The coefficients of the balanced chemical equation for a reaction indicate the relative numbers of
moles of each reactant that combine during the process, as well as the number of moles of each
product formed.
3.
Although we define mass as the “amount of matter in a substance,” the units in which we measure
mass are a human invention. Atoms and molecules react on an individual particle-by-particle
basis, and we have to count individual particles when doing chemical calculations.
4.
Balanced chemical equations tell us in what molar ratios substances combine to form products,
not in what mass proportions they combine.
5.
a.
PCl3(l) + 3H2O(l) → H3PO3(aq) + 3HCl(g)
One molecule of liquid phosphorus trichloride reacts with three molecules of liquid
water, producing one molecule of aqueous phosphorous acid and molecules of gaseous
hydrogen chloride. One mole of phosphorus trichloride reacts with three moles of water
to produce one mole of phosphorous acid and three moles of hydrogen chloride.
b.
2XeF2(g) + 2H2O(l) → 2Xe(g) + 4HF(g) + O2(g)
Two molecules of gaseous xenon difluoride react with two molecules of liquid water,
producing two gaseous xenon atoms, four molecules of gaseous hydrogen fluoride, and
one molecule of oxygen gas. Two moles of xenon difluoride reacts with two moles of
water, to produce two moles of xenon, four moles of hydrogen fluoride, and one mole of
oxygen.
c.
S(s) + 6HNO3(aq) → H2SO4(aq) + 2H2O(l) + 6NO2(g)
One sulfur atom reacts with six molecules of aqueous nitric acid, producing one molecule
of aqueous sulfuric acid, two molecules of water, and six molecules of nitrogen dioxide
gas. One mole of sulfur reacts with six moles of nitric acid, to produce one mole of
sulfuric acid, two moles of water, and six moles of nitrogen dioxide.
d.
2NaHSO3(s) → Na2SO3(s) + SO2(g) + H2O(l)
Two formula units of solid sodium hydrogen sulfite react to produce one formula unit of
solid sodium sulfite, one molecule of gaseous sulfur dioxide, and one molecule of liquid
water. Two moles of sodium hydrogen sulfite react to produce one mole of sodium
sulfite, one mole of sulfur dioxide, and one mole of water.
6.
a.
(NH4)2CO3(s) → 2NH3(g) + CO2(g) + H2O(g)
One formula unit of solid ammonium carbonate decomposes to produce two molecules of
ammonia gas, one molecule of carbon dioxide gas, and one molecule of water vapor.
163
Chapter 9:
Chemical Quantities
One mole of solid ammonium carbonate decomposes into two moles of gaseous
ammonia, one mole of carbon dioxide gas, and one mole of water vapor.
b.
6Mg(s) + P4(s) → 2Mg3P2(s)
Six atoms of magnesium metal react with one molecule of solid phosphorus (P4) to make
two formula units of solid magnesium phosphide. Six moles of magnesium metal react
with one mole of phosphorus solid (P4) to produce two moles of solid magnesium
phosphide.
c.
4Si(s) + S8(s) → 2Si2S4(l)
Four atoms of solid silicon react with one molecule of solid sulfur (S8) to form two
molecules of liquid disilicon tetrasulfide. Four moles of solid silicon react with one mole
of solid sulfur (S8) to form two moles of liquid disilicon tetrasulfide.
d.
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
One molecule of liquid ethanol burns with three molecules of oxygen gas to produce two
molecules of carbon dioxide gas and three molecules of water vapor. One mole of liquid
ethanol burns with three moles of oxygen gas to produce two moles of gaseous carbon
dioxide and three moles of water vapor.
7.
False. The coefficients of the balanced chemical equation represent the ratios on a mole basis by
which potassium hydroxide combines with sulfur dioxide.
8.
Balanced chemical equations tell us in what molar ratios substances combine to form products,
not in what mass proportions they combine. How could 2 grams of reactant produce a total of 3
grams of products?
9.
4Al(s) + 3O2(g) → 2Al2O3(s)
For converting from a given number of moles of aluminum metal to the number of moles of
oxygen needed for reaction, the correct mole ratio is
⎛ 3 mol O 2 ⎞
⎜
⎟
⎝ 4 mol Al ⎠
For converting from a given number of moles of aluminum metal to the number of moles of
product produced, the mole ratio is
⎛ 2 mol Al2 O3 ⎞
⎜ 4 mol Al ⎟
⎝
⎠
10.
Fe2O3(s) + 3H2SO4(aq) → Fe2(SO4)3(s) + 3H2O(l)
For converting from a given number of moles of iron(III) oxide to the number of moles of
sulfuric acid required, the mole ratio is
⎛ 3 mol H 2SO 4 ⎞
⎜
⎟
⎝ 1 mol Fe 2 O3 ⎠
For a given number of moles of iron(III) oxide reacting completely, the mole ratios used to
calculate the number of moles of each product are
164
Chapter 9:
⎛ 1 mol Fe 2 (SO 4 )3 ⎞
For Fe2(SO4)3: ⎜
⎟
⎝ 1 mol Fe 2 O3 ⎠
11.
a.
b.
c.
d.
12.
a.
b.
c.
⎛ 3 mol H 2 O ⎞
For H2O: ⎜
⎟
⎝ 1 mol Fe 2 O3 ⎠
CO2(g) + 4H2(g) → CH4(g) + 2H2O(l)
0.500 mol CO2 ×
1 mol CH 4
= 0.500 mol CH4
1 mol CO 2
0.500 mol CO2 ×
2 mol H 2 O
= 1.00 mol H2O
1 mol CO 2
BaCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Ba(NO3)2(aq)
0.500 mol BaCl2 ×
2 mol AgCl
= 1.00 mol AgCl
1 mol BaCl2
0.500 mol BaCl2 ×
1 mol Ba(NO3 ) 2
= 0.500 mol Ba(NO3)2
1 mol BaCl2
C3H8(g) + 5O2(g) → 4H2O(l) + 3CO2(g)
0.500 mol C3H8 ×
4 mol H 2 O
= 2.00 mol H2O
1 mol C3 H8
0.500 mol C3H8 ×
3 mol CO 2
= 1.50 mol CO2
1 mol C3 H8
3H2SO4(aq) + 2Fe(s) → Fe2(SO4)3(aq) + 3H2(g)
0.500 mol H2SO4 ×
1 mol Fe 2 (SO 4 )3
= 0.167 mol Fe2(SO4)3
3 mol H 2SO 4
0.500 mol H2SO4 ×
3 mol H 2
= 0.500 mol H2
3 mol H 2SO 4
4Bi(s) + 3O2(g) → 2Bi2O3(s)
0.250 mol Bi ×
2 mol Bi 2 O3
= 0.125 mol Bi2O3
4 mol Bi
SnO2(s) + 2H2(g) → Sn(s) + 2H2O(g)
0.250 mol SnO2 ×
1 mol Sn
= 0.250 mol Sn
1 mol SnO 2
0.250 mol SnO2 ×
2 mol H 2 O
= 0.500 mol H2O
1 mol SnO 2
SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(g)
0.250 mol SiCl4 ×
Chemical Quantities
1 mol SiO 2
= 0.250 mol SiO2
1 mol SiCl4
165
Chapter 9:
Chemical Quantities
0.250 mol SiCl4 ×
d.
2N2(g) + 5O2(g) + 2H2O(l) → 4HNO3(aq)
0.250 mol N2 ×
13.
a.
4 mol HCl
= 1.00 mol HCl
1 mol SiCl4
4 mol HNO3
= 0.500 mol HNO3
2 mol N 2
AgNO3(aq) + LiOH(aq) → AgOH(s) + LiNO3(aq)
molar masses: AgOH, 124.91 g; LiNO3, 68.95 g
0.125 mol AgNO3 ×
124.91 g AgOH
= 15.6 g AgOH
1 mol AgOH
0.125 mol AgOH ×
b.
1 mol AgOH
= 0.125 mol AgOH
1 mol AgNO3
0.125 mol AgNO3 ×
1 mol LiNO3
= 0.125 mol LiNO3
1 mol AgNO3
0.125 mol LiNO3 ×
68.95 g LiNO3
= 8.62 g LiNO3
1 mol LiNO3
Al2(SO4)3(aq) + 3CaCl2(aq) → 2AlCl3(aq) + 3CaSO4(s)
molar masses: AlCl3, 133.33 g; CaSO4, 136.15 g
0.125 mol Al2(SO4)3 ×
0.250 mol AlCl3 ×
133.33 g AlCl3
= 33.3 g AlCl3
1 mol AlCl3
0.125 mol Al2(SO4)3 ×
3 mol CaSO 4
= 0.375 mol CaSO4
1 mol Al2 (SO 4 )3
136.15 g CaSO4
= 51.1 g CaSO4
1 mol CaSO 4
0.375 mol CaSO4 ×
c.
2 mol AlCl3
= 0.250 mol AlCl3
1 mol Al2 (SO 4 )3
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
molar masses: CaCl2, 110.98 g; CO2, 44.01 g; H2O, 18.02 g
0.125 mol CaCO3 ×
0.125 mol CaCl2 ×
1 mol CaCl2
= 0.125 mol CaCl2
1 mol CaCO3
110.98 g CaCl2
= 13.9 g CaCl2
1 mol CaCl2
0.125 mol CaCO3 ×
1 mol CO 2
= 0.125 mol CO2
1 mol CaCO3
166
Chapter 9:
0.125 mol CO2 ×
44.01 g CO 2
= 5.50 g CO2
1 mol CO 2
1 mol H 2 O
= 0.125 mol H2O
1 mol CaCO3
0.125 mol CaCO3 ×
0.125 mol H2O ×
d.
18.02 g H 2 O
= 2.25 g H2O
1 mol H 2 O
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
molar masses: CO2, 44.01 g; H2O, 18.02 g
0.125 mol C4H10 ×
0.500 mol CO2 ×
44.01 g CO 2
= 22.0 g CO2
1 mol CO 2
0.125 mol C4H10 ×
0.625 mol H2O ×
14.
a.
8 mol CO 2
= 0.500 mol CO2
2 mol C4 H10
10 mol H 2 O
= 0.625 mol H2O
2 mol C4 H10
18.02 g H 2 O
= 11.3 g H2O
1 mol H 2 O
C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l)
molar masses: CO2, 44.01 g; H2O, 18.02 g
0.750 mol C5H12 ×
3.75 mol CO2 ×
44.01 g CO 2
= 165 g CO2
1 mol CO 2
0.750 mol C5H12 ×
4.50 mol H2O ×
b.
5 mol CO 2
= 3.75 mol CO2
1 mol C5 H12
6 mol H 2 O
= 4.50 mol H2O
1 mol C5 H12
18.02 g H 2 O
= 81.1 g H2O
1 mol H 2 O
2CH3OH(l) + 3O2(g) → 4H2O(l) + 2CO2(g)
molar masses: H2O, 18.02 g; CO2, 44.01 g
0.750 mol CH3OH ×
1.50 mol H2O ×
4 mol H 2 O
= 1.50 mol H2O
2 mol CH 3OH
18.02 g H 2 O
= 27.0 g H2O
1 mol H 2 O
0.750 mol CH3OH ×
2 mol CO 2
= 0.750 mol CO2
2 mol CH 3OH
167
Chemical Quantities
Chapter 9:
Chemical Quantities
0.750 mol CO2 ×
c.
44.01 g CO 2
= 33.0 g CO2
1 mol CO 2
Ba(OH)2(aq) + H3PO4(aq) → BaHPO4(s) + 2H2O(l)
molar masses: BaHPO4, 233.3 g; H2O, 18.02 g
0.750 mol Ba(OH)2 ×
1 mol BaHPO 4
= 0.750 mol BaHPO4
1 mol Ba(OH) 2
0.750 mol BaHPO4 ×
233.3 g BaHPO 4
= 175 g BaHPO4
1 mol BaHPO 4
0.750 mol Ba(OH)2 ×
2 mol H 2 O
= 1.50 mol H2O
1 mol Ba(OH) 2
1.50 mol H2O ×
d.
18.02 g H 2 O
= 27.0 g H2O
1 mol H 2 O
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)
molar masses: C2H5OH, 46.07 g; CO2, 44.01 g
0.750 mol C6H12O6 ×
1.50 mol C2H5OH ×
0.750 mol C6H12O6 ×
1.50 mol CO2 ×
15.
a.
44.01 g CO 2
= 66.0 g CO2
1 mol CO 2
2 mol KI
= 0.550 mol KI
1 mol Cl2
1 mol P4
= 0.0458 mol P4
6 mol Co
Zn(s) + 2HNO3(aq) → Zn(NO3)2(aq) + H2(g)
0.275 mol Zn ×
d.
2 mol CO 2
= 1.50 mol CO2
1 mol C6 H12 O6
6Co(s) + P4(s) → 2Co3P2(s)
0.275 mol Co ×
c.
46.07 g C 2 H 5 OH
= 69.1 g C2H5OH
1 mol C2 H 5 OH
Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(s)
0.275 mol Cl2 ×
b.
2 mol C 2 H 5 OH
= 1.50 mol C2H5OH
1 mol C6 H12 O6
2 mol HNO3
= 0.550 mol HNO3
1 mol Zn
C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(g)
0.275 mol C5H12 ×
8 mol O 2
= 2.20 mol O2
1 mol C5 H12
168
Chapter 9:
16.
Chemical Quantities
Before doing the calculations, the equations must be balanced.
a.
4KO2(s) + 2H2O(l) → 3O2(g) + 4KOH(s)
3 mol O 2
= 0.469 mol O2
4 mol KOH
0.625 mol KOH ×
b.
SeO2(g) + 2H2Se(g) → 3Se(s) + 2H2O(g)
0.625 mol H2O ×
c.
2CH3CH2OH(l) + O2(g) → 2CH3CHO(aq) + 2H2O(l)
0.625 mol H2O ×
d.
3 mol Se
= 0.938 mol Se
2 mol H 2 O
2 mol CH 3CHO
= 0.625 mol CH3CHO
2 mol H 2 O
Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s)
0.625 mol Al2O3 ×
2 mol Fe
= 1.25 mol Fe
1 mol Al2 O3
17.
the molar mass of the substance
18.
Stoichiometry is the process of using a chemical equation to calculate the relative masses of
reactants and products involved in a reaction.
19.
a.
molar mass Si = 28.09 g
4.15 g Si ×
b.
1 mol Si
= 0.148 mol Si
28.09 g Si
molar mass AuCl3 = 303.4 g
2.72 mg AuCl3 ×
c.
molar mass S = 32.07 g
1.05 kg S ×
d.
1 mol AuCl3
1g
×
= 8.97 × 10–6 mol AuCl3
1000 mg
303.4 g AuCl3
1000 g
1 mol S
×
= 32.7 mol S
1 kg
32.07 g S
molar mass FeCl3 = 162.20 g
0.000901 g FeCl3 ×
e.
1 mol FeCl3
= 5.55 × 10–6 mol FeCl3
162.2 g FeCl3
molar mass MgO = 40.31 g
5.62 × 103 g MgO ×
1 mol MgO
= 139 mol MgO
40.31 g MgO
169
Chapter 9:
20.
a.
Chemical Quantities
molar mass Ar = 39.95 g
1g
1 mol Ar
×
= 1.81 × 10–3 mol Ar
1000 mg
39.95 g Ar
72.4 mg Ar ×
b.
molar mass CS2 = 76.15 g
52.7 g CS2 ×
c.
molar mass Fe = 55.85 g
784 kg Fe ×
d.
1 mol CS2
= 0.692 mol CS2
76.15 g CS2
1000 g
1 mol Fe
×
= 1.40 × 104 mol Fe
1 kg
55.85 g Fe
molar mass CaCl2 = 110.98 g
0.00104 g CaCl2 ×
e.
molar mass NiS = 90.76 g
1.26 × 103 g NiS ×
21.
a.
1 mol NiS
= 13.9 mol NiS
90.76 g NiS
molar mass Ge = 72.59 g
2.17 mol Ge ×
b.
1 mol CaCl2
= 9.37 × 10–6 mol CaCl2
110.98 g CaCl2
72.59 g Ge
= 158 g Ge
1 mol Ge
molar mass PbCl2 = 278.1 g;
0.00424 mol PbCl2 ×
c.
278.1 g PbCl2
= 1.18 g PbCl2
1 mol PbCl2
molar mass NH3 = 17.03 g
0.0971 mol NH3 ×
d.
17.03 g NH 3
= 1.65 g NH3
1 mol NH 3
molar mass C6H14 = 86.17 g
4.26 × 103 mol C6H14 ×
e.
a.
86.17 g C6 H14
= 3.67 × 105 g C6H14
1 mol C6 H14
molar mass ICl = 162.35 g
1.71 mol ICl ×
22.
4.24 millimol = 0.00424 mol
162.35 g ICl
= 278 g ICl
1 mol ICl
molar mass of C3H8 = 44.09 g
2.23 mol C3H8 ×
44.09 g C3 H8
= 98.3 g C3H8
1 mol C3 H8
170
Chapter 9:
b.
molar mass of Ar = 39.95 g; 9.03 millimol = 0.00903 mol
0.00903 mol Ar ×
c.
39.95 g Ar
= 0.361 g Ar
1 mol Ar
molar mass of SiO2 = 60.09 g
5.91 × 106 mol SiO2 ×
d.
molar mass of CuCl2 = 134.45 g
0.000104 mol CuCl2 ×
e.
99.00 g CuCl
= 0.0103 g CuCl
1 mol CuCl
Before any calculations are done, the equations must be balanced.
a.
2Co(s) + 3F2(g) → 2CoF3(s)
0.413 mol Co ×
b.
c.
3 mol H 2SO 4
= 0.620 mol H2SO4
2 mol Al
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)
0.413 mol K ×
d.
3 mol F2
= 0.620 mol F2
2 mol Co
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
0.413 mol Al ×
2 mol H 2 O
= 0.413 mol H2O
2 mol K
4Cu(s) + O2(g) → 2Cu2O(s)
0.413 mol Cu ×
24.
134.45 g CuCl2
= 0.0140 g CuCl2
1 mol CuCl2
molar mass of CuCl = 99.00 g
0.000104 mol CuCl ×
23.
60.09 g SiO 2
= 3.55 × 108 g SiO2
1 mol SiO 2
1 mol O 2
= 0.103 mol O2
4 mol Cu
Before any calculations are done, the equations must be balanced.
a.
2Al(s) + 3Br2(l) → 2AlBr3(s)
molar mass Al = 26.98 g
0.557 g Al ×
b.
1 mol Al 3 mol Br2
×
= 0.0310 mol Br2
26.98 g Al 2 mol Al
Hg(s) + 2HClO4(aq) → Hg(ClO4)2(aq) + H2(g)
molar mass Hg = 200.6 g
0.557 g Hg ×
1 mol Hg 2 mol HClO 4
×
= 0.00555 mol HClO4
200.6 g
1 mol Hg
171
Chemical Quantities
Chapter 9:
c.
Chemical Quantities
3K(s) + P(s) → K3P(s)
molar mass K = 39.10 g
0.557 g K ×
d.
1 mol K
1 mol P
×
= 0.00475 mol P
39.10 g K 3 mol K
CH4(g) + 4Cl2(g) → CCl4(l) + 4HCl(g)
molar mass CH4 = 16.04 g
0.557 g CH4 ×
25.
1 mol CH 4
4 mol Cl2
×
= 0.139 mol Cl2
16.04 g CH 4 1 mol CH 4
Before any calculations are done, the equations must be balanced.
a.
TiBr4(g) + 2H2(g) → Ti(s) + 4HBr(g)
molar mass H2 = 2.016 g; molar mass Ti = 47.90 g; molar mass of HBr = 80.91 g
12.5 g H2 ×
1 mol H 2
= 6.20 mol H2
2.016 g H 2
6.20 mol H2 ×
1 mol Ti
= 3.10 mol Ti
2 mol H 2
3.10 mol Ti ×
47.90 g Ti
= 148 g Ti
1 mol Ti
6.20 mol H2 ×
4 mol HBr
= 12.4 mol HBr
2 mol H 2
12.4 mol HBr ×
b.
80.91 g HBr
= 1.00 × 103 g HBr
1 mol HBr
3SiH4(g) + 4NH3(g) → Si3N4(s) + 12H2(g)
molar mass SiH4 = 32.12 g; molar mass Si3N4 = 140.3 g; molar mass H2 = 2.016 g
12.5 g SiH4 ×
1 mol SiH 4
= 0.389 mol SiH4
32.12 g SiH 4
0.389 mol SiH4 ×
1 mol Si3 N 4
= 0.130 mol Si3N4
3 mol SiH 4
0.130 mol Si3N4 ×
140.3 g Si3 N 4
= 18.2 g Si3N4
1 mol Si3 N 4
0.389 mol SiH4 ×
12 mol H 2
= 1.56 mol H2
3 mol SiH 4
1.56 mol H2 ×
2.016 g H 2
= 3.14 g H2
1 mol H 2
172
Chapter 9:
c.
Chemical Quantities
2NO(g) + 2H2(g) → N2(g) + 2H2O(l)
molar mass H2 = 2.016 g; molar mass N2 = 28.02 g; molar mass H2O = 18.02 g
12.5 g H2 ×
1 mol H 2
= 6.20 mol H2
2.016 g H 2
6.20 mol H2 ×
1 mol N 2
= 3.10 mol N2
2 mol H 2
3.10 mol N2 ×
28.02 g N 2
= 86.9 g N2
1 mol N 2
6.20 mol H2 ×
2 mol H 2 O
= 6.20 mol H2O
2 mol H 2
6.20 mol H2O ×
d.
18.02 g H 2 O
= 112 g H2O
1 mol H 2 O
Cu2S(s) → 2Cu(s) + S(g)
molar mass Cu2S = 159.2 g; molar mass Cu = 63.55 g; molar mass S = 32.07 g
12.5 g Cu2S ×
1 mol Cu 2S
= 0.0785 mol Cu2S
159.2 g Cu 2S
0.0785 mol Cu2S ×
0.157 mol Cu ×
63.55 g Cu
= 9.98 g Cu
1 mol Cu
0.0785 mol Cu2S ×
0.0785 mol S ×
26.
a.
2 mol Cu
= 0.157 mol Cu
1 mol Cu 2S
1 mol S
= 0.0785 mol S
1 mol Cu 2S
32.07 g S
= 2.52 g S
1 mol S
2BCl3(s) + 3H2(g) → 2B(s) + 6HCl(g)
molar masses: BCl3, 117.16 g; B, 10.81 g; HCl, 36.46 g
15.0 g BCl3 ×
b.
1 mol BCl3
= 0.128 mol BCl3
117.16 g BCl3
0.128 mol BCl3 ×
2 mol B
10.81 g B
×
= 1.38 g B
2 mol BCl3 1 mol B
0.128 mol BCl3 ×
6 mol HCl 36.46 g HCl
×
= 14.0 g HCl
2 mol BCl3 1 mol HCl
2Cu2S(s) + 3O2(g) → 2Cu2O(s) + 2SO2(g)
molar masses: Cu2S, 159.17 g; Cu2O, 143.1 g; SO2, 64.07 g
173
Chapter 9:
Chemical Quantities
15.0 g Cu2S ×
c.
1 mol Cu 2S
= 0.09424 mol Cu2S
159.17 g Cu 2S
0.09424 mol Cu2S ×
2 mol Cu 2 O 143.1 g Cu 2 O
×
= 13.5 g Cu2O
2 mol Cu 2S 1 mol Cu 2 O
0.09424 mol Cu2S ×
2 mol SO 2 64.07 g SO 2
×
= 6.04 g SO2
2 mol Cu 2S 1 mol SO 2
2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2(g)
molar masses: Cu2S, 159.17 g; Cu, 63.55 g; SO2, 64.07 g
15.0 g Cu2S ×
d.
1 mol Cu 2S
= 0.09424 mol Cu2S
159.17 g Cu 2S
0.09424 mol Cu2S ×
6 mol Cu 63.55 g Cu
×
= 35.9 g Cu
1 mol Cu 2S 1 mol Cu
0.09424 mol Cu2S ×
1 mol SO 2 64.07 g SO 2
×
= 6.04 g SO2
1 mol Cu 2S 1 mol SO 2
CaCO3(s) + SiO2(s) → CaSiO3(s) + CO2(g)
molar masses: SiO2, 60.09 g; CaSiO3, 116.17 g; CO2, 44.01 g
15.0 g SiO2 ×
27.
1 mol SiO 2
= 0.2496 mol SiO2
60.09 g SiO 2
0.2496 mol SiO2 ×
1 mol CaSiO3 116.17 g CaSiO3
×
= 29.0 g CaSiO3
1 mol SiO 2
1 mol CaSiO3
0.2496 mol SiO2 ×
1 mol CO 2 44.01 g CO 2
×
= 11.0 g CO2
1 mol SiO 2 1 mol CO 2
The equation must first be balanced.
(NH4)2CO3(s) → 2NH3(g) + CO2(g) + H2O(g
molar masses: (NH4)2CO3, 96.09 g; NH3, 17.03 g
1.25 g (NH4)2CO3 ×
1 mol ( NH 4 )2 CO3
96.09 g ( NH 4 )2 CO3
0.0130 mol (NH4)2CO3 ×
0.0260 mol NH3 ×
28.
= 0.0130 mol (NH4)2CO3
2 mol NH 3
= 0.0260 mol NH3
1 mol ( NH 4 )2 CO3
17.03 g NH 3
= 0.443 g NH3
1 mol NH 3
The balanced equation for the reaction is:
CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)
174
Chapter 9:
molar masses: CaC2, 64.10 g; C2H2, 26.04 g
3.75 g CaC2 ×
29.
1 mol CaC2
= 0.0585 mol CaC2
64.10 g CaC 2
0.0585 mol CaC2×
1 mol C2 H 2
= 0.0585 mol C2H2
1 mol CaC 2
0.0585 mol C2H2 ×
26.04 g C2 H 2
= 1.52 g C2H2
1 mol C2 H 2
molar masses: C, 12.01 g; CO, 28.01 g; CO2, 44.01 g
5.00 g C ×
1 mol C
= 0.4163 mol C
12.01 g C
carbon dioxide: C(s) + O2(g) → CO2(g)
0.4163 mol C ×
1 mol CO 2
= 0.4163 mol CO2
1 mol C
0.4163 mol CO2 ×
44.01 g CO 2
= 18.3 g CO2
1 mol CO 2
carbon monoxide: 2C(s) + O2(g) → 2CO(g)
0.4163 mol C ×
2 mol CO
= 0.4163 mol CO
2 mol C
0.4163 mol CO ×
30.
28.01 g CO
= 11.7 g CO
1 mol CO
2NaHCO3(s) → Na2CO3(s) + H2O(g) + CO2(g)
molar masses: NaHCO3, 84.01 g; Na2CO3, 106.0 g
1.52 g NaHCO3 ×
31.
1 mol NaHCO3
= 0.01809 mol NaHCO3
84.01 g NaHCO3
0.01809 mol NaHCO3 ×
1 mol Na 2 CO3
= 0.009047 mol Na2CO3
2 mol NaHCO3
0.009047 mol Na2CO3 ×
106.0 g Na 2 CO3
= 0.959 g Na2CO3
1 mol Na 2 CO3
2Fe(s) + 3Cl2(g) → 2FeCl3(s)
millimolar masses: iron, 55.85 mg; FeCl3, 162.2 mg
15.5 mg Fe ×
1 mmol Fe
= 0.2775 mmol Fe
55.85 mg Fe
175
Chemical Quantities
Chapter 9:
Chemical Quantities
0.2775 mmol Fe ×
2 mmol FeCl3
= 0.2775 mmol FeCl3
2 mmol Fe
0.2775 mmol FeCl3 ×
32.
162.2 mg FeCl3
= 45.0 mg FeCl3
1 mmol FeCl3
C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)
molar masses: C6H12O6, 180.2 g; C2H5OH, 46.07 g
5.25 g C6H12O6 ×
1 mol C6 H12 O6
= 0.02913 mol C6 H12 O6
180.2 g C6 H12 O6
0.02913 mol C6 H12 O6 ×
0.5286 mol C2H5OH ×
33.
2 mol C 2 H 5 OH
= 0.5826 mol C2H5OH
1 mol C6 H12 O6
46.07 g C2 H5 OH
= 2.68 g ethyl alcohol
1 mol C2 H 5OH
H2SO3(aq) → H2O(l) + SO2(g)
molar masses: H2SO3, 82.09 g; SO2, 64.07 g
4.25 g H2SO3 ×
1 mol H 2SO3
= 0.05177 mol H2SO3
82.09 g H 2SO3
0.05177 mol H2SO3 ×
0.05177 mol SO2 ×
34.
1 mol SO 2
= 0.05177 mol SO2
1 mol H 2SO3
64.07 g SO 2
= 3.32 g SO2
1 mol SO 2
NH4Cl(s) + NaOH(s) → NH3(g) + NaCl(s) + H2O(g)
molar masses: NH4Cl, 53.49 g; NH3, 17.03 g
1.39 g NH4Cl ×
1 mol NH 4 Cl
= 0.02599 mol NH4Cl
53.49 g NH 4 Cl
0.02599 mol NH4Cl ×
0.02599 mol NH3 ×
35.
1 mol NH 3
= 0.02599 mol NH3
1 mol NH 4 Cl
17.03 g NH 3
= 0.443 g NH3
1 mol NH3
P4(s) + 5O2(g) → 2P2O5(s)
molar masses: P4, 123.88 g; O2, 32.00 g
4.95 g P4 ×
1 mol P4
= 0.03996 mol P4
123.88 g P4
176
Chapter 9:
5 mol O 2
= 0.1998 mol O2
1 mol P4
0.03996 mol P4 ×
32.00 g O 2
= 6.39 g O2
1 mol O 2
0.1998 mol O2 ×
36.
4HgS(s) + 4CaO(s) → 4Hg(l) + 3CaS(s) + CaSO4(s)
molar masses: HgS, 232.7 g Hg, 200.6 g; 10.0 kg = 1.00 × 104 g
1.00 × 104 g HgS ×
4 mol Hg
= 42.97 mol Hg
4 mol HgS
42.97 mol HgS ×
42.97 mol Hg ×
37.
1 mol HgS
= 42.97 mol HgS
232.7 g HgS
200.6 g Hg
= 8.62 × 103 g Hg = 8.62 kg Hg
1 mol Hg
2NH4NO3(s) → 2N2(g) + O2(g) + 4H2O(g)
molar masses: NH4NO3, 80.05 g; N2, 28.02 g; 02, 32.00 g; H2O, 18.02 g
1 mol NH 4 NO3
= 0.0156 mol NH4NO3
80.05 g NH 4 NO3
1.25 g NH4NO3 ×
0.0156 mol NH4NO3 ×
0.0156 mol N2 ×
2 mol N 2
= 0.0156 mol N2
2 mol NH 4 NO3
28.02 g N 2
= 0.437 g N2
1 mol N 2
0.0156 mol NH4NO3 ×
0.00780 mol O2 ×
32.00 g O 2
= 0.250 g O2
1 mol O 2
0.0156 mol NH4NO3 ×
0.0312 mol H2O ×
1 mol O 2
= 0.00780 mol O2
2 mol NH 4 NO3
4 mol H 2 O
= 0.0312 mol H2O
2 mol NH 4 NO3
18.02 g H 2 O
= 0.562 g H2O
1 mol H 2 O
As a check, note that 0.437 g + 0.250 g + 0.562 g = 1.249 g = 1.25 g.
38.
C12H22O11(s) → 12C(s) + 11H2O(g)
molar masses: C12H22O11, 342.3 g; C, 12.01
1.19 g C12H22O11 ×
1 mol C12 H 22 O11
= 3.476 × 10–3 mol C12 H 22 O11
342.3 g C12 H 22 O11
177
Chemical Quantities
Chapter 9:
Chemical Quantities
3.476 × 10–3 mol C12 H 22 O11 ×
0.04172 mol C ×
39.
12 mol C
= 0.04172 mol C
1 mol C12 H 22 O11
12.01 g C
= 0.501 g C
1 mol C
SOCl2(l) + H2O(l) → SO2(g) + 2HCl(g)
molar masses: SOCl2, 119.0 g; H2O, 18.02 g
35.0 g SOCl2 ×
1 mol SOCl2
= 0.294 mol SOCl2
119.0 g SOCl2
0.294 mol SOCl2 ×
0.294 mol H2O ×
40.
1 mol H 2 O
= 0.294 mol H2O
1 mol SOCl2
18.02 g H 2 O
= 5.30 g H2O
1 mol H 2 O
The balanced equation is:
2C8H18 + 25O2 → 16CO2 + 18H2O
molar masses: C8H18, 114.22 g; CO2, 44.01 g
453.59 g CO2 ×
1 lb of CO2 = 453.59 g CO2
1 mol CO 2
= 10.31 mol CO2
44.01 g CO 2
From the balanced chemical equation, we can calculate the number of moles and number of
grams of pure octane that would be required to produce 10.31 mol CO2.
10.31 mol CO2 ×
2 mol C8 H18
= 1.288 mol C8H18
16 mol CO 2
1.288 mol C8H18 ×
114.22 g C8 H18
= 147.2 g C8H18
1 mol C8 H18
From the density of C8H18 we can calculate the volume of 147.2 g C8H18.
147.2 g C8H18 ×
1 mL C8 H18
= 196.3 mL C8H18 (2.0 × 102 mL to two significant figures)
0.75 g C8 H18
From the preceding, we know that to travel 1 mile, we need approximately 200 mL of octane
1 mi
1000 mL 3.7854 L
×
×
= approximately 19 mi/gal
196.3 mL
1L
1 gal
41.
To determine the limiting reactant, first calculate the number of moles of each reactant present.
Then determine how these numbers of moles correspond to the stoichiometric ratio indicated by
the balanced chemical equation for the reaction. Specific answer depends on student response.
42.
To determine the limiting reactant, first calculate the number of moles of each reactant present.
Then determine how these numbers of moles correspond to the stoichiometric ratio indicated by
the balanced chemical equation for the reaction.
178
Chapter 9:
Chemical Quantities
43.
The theoretical yield of a reaction represents the stoichiometric amount of product that should
form if the limiting reactant for the process is completely consumed.
44.
A reactant is present in excess if there is more of that reactant present than is needed to combine
with the limiting reactant for the process. By definition, the limiting reactant cannot be present in
excess. An excess of any reactant does not affect the theoretical yield for a process: the
theoretical yield is determined by the limiting reactant.
45.
a.
Na2B4O7(s) + H2SO4(aq) + 5H2O(l) → 4H3BO3(s) + Na2SO4(aq)
molar masses: Na2B4O7, 201.2 g; H2SO4, 98.09 g; H2O, 18.02 g
5.00 g Na2B4O7 ×
5.00 g H2SO4 ×
5.00 g H2O ×
1 mol
= 0.0249 mol Na2B4O7
201.2 g
1 mol
= 0.0510 mol H2SO4
98.09 g
1 mol
= 0.277 mol H2O
18.02 g
Na2B4O7 is the limiting reactant.
mol H2SO4 remaining unreacted = 0.0510 – 0.0249 = 0.0261 mol
mol H2O remaining unreacted = 0.277 – 5(0.0249) = 0.153 mol
mass of H2SO4 remaining = 0.0261 mol ×
mass of H2O remaining = 0.153 mol ×
b.
98.09 g
= 2.56 g H2SO4
1 mol
18.02 g
= 2.76 g H2O
1 mol
CaC2(s) + 2H2O(l) → Ca(OH)2(s) + C2H2(g)
molar masses: CaC2, 64.10 g; H2O, 18.02 g
5.00 g CaC2 ×
5.00 g H2O ×
1 mol
= 0.0780 mol CaC2
64.10 g
1 mol
= 0.277 mol H2O
18.02 g
CaC2 is the limiting reactant; water is present in excess.
mol of H2O remaining = 0.277 – 2(0.0780) = 0.121 mol H2O
mass of H2O remaining = 0.121 mol ×
c.
18.02 g
= 2.18 g H2O
1 mol
2NaCl(s) + H2SO4(l) → 2HCl(g) + Na2SO4(s)
molar masses: NaCl, 58.44 g; H2SO4, 98.09 g
5.00 g NaCl ×
1 mol
= 0.0856 mol NaCl
58.44 g
179
Chapter 9:
Chemical Quantities
5.00 g H2SO4 ×
1 mol
= 0.0510 mol H2SO4
98.09 g
NaCl is the limiting reactant; H2SO4 is present in excess.
mol H2SO4 that reacts = 0.5(0.0856) = 0.0428 mol H2SO4
mol H2SO4 remaining = 0.0510 – 0.0428 = 0.0082 mol
mass of H2SO4 remaining = 0.0082 mol ×
d.
98.09 g
= 0.80 g H2SO4
1 mol
SiO2(s) + 2C(s) → Si(l) + 2CO(g)
molar masses: SiO2, 60.09 g; C, 12.01 g
5.00 g SiO2 ×
5.00 g C ×
1 mol
= 0.0832 mol SiO2
60.09 g
1 mol
= 0.416 mol C
12.01 g
SiO2 is the limiting reactant; C is present in excess.
mol C remaining = 0.416 – 2(0.0832) = 0.250 mol
mass of C remaining = 0.250 mol ×
46.
a.
12.01 g
= 3.00 g C
1 mol
S(s) + 2H2SO4(aq) → 3SO2(g) + 2H2O(l)
Molar masses: S, 32.07 g; H2SO4, 98.09 g; SO2, 64.07 g; H2O, 18.02 g
5.00 g S ×
1 mol
= 0.1559 mol S
32.07 g
5.00 g H2SO4 ×
1 mol
= 0.05097 mol H2SO4
98.09 g
According to the balanced chemical equation, we would need twice as much sulfuric acid
as sulfur for complete reaction of both reactants. We clearly have much less sulfuric acid
present than sulfur: sulfuric acid is the limiting reactant. The calculation of the masses of
products produced is based on the number of moles of the sulfuric acid.
b.
0.05097 mol H2SO4 ×
3 mol SO 2
64.07 g SO 2
×
= 4.90 g SO2
2 mol H 2SO 4 1 mol SO 2
0.05097 mol H2SO4 ×
2 mol H 2 O 18.02 g H 2 O
×
= 0.918 g H2O
2 mol H 2SO 4 1 mol H 2 O
MnO2(s) + 2H2SO4(aq) → Mn(SO4)2 + 2H2O(l)
molar masses: MnO2, 86.94 g; H2SO4 98.09 g; Mn(SO4)2, 247.1 g; H2O, 18.02 g
5.00 g MnO2 ×
1 mol
= 0.05751 mol MnO2
86.94 g
180
Chapter 9:
5.00 g H2SO4 ×
Chemical Quantities
1 mol
= 0.05097 mol H2SO4
98.09 g
According to the balanced chemical equation, we would need twice as much sulfuric acid
as manganese(IV) oxide for complete reaction of both reactants. We do not have this
much sulfuric acid, so sulfuric acid must be the limiting reactant. The amount of each
product produced will be based on the sulfuric acid reacting completely.
c.
0.05097 mol H2SO4 ×
1 mol Mn(SO 4 ) 2 247.1 g Mn(SO 4 ) 2
×
= 6.30 g Mn(SO4)2
2 mol H2SO4
1 mol Mn(SO 4 ) 2
0.05097 mol H2SO4 ×
2 mol H 2 O 18.02 g H 2 O
×
= 0.918 g H2O
2 mol H 2SO 4 1 mol H 2 O
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(l)
Molar masses: H2S, 34.09 g; O2, 32.00 g; SO2, 64.07 g; H2O, 18.02 g
5.00 g H2S ×
5.00 g O2 ×
1 mol
= 0.1467 mol H2S
34.09 g
1 mol
= 0.1563 mol O2
32.00 g
According to the balanced equation, we would need 1.5 times as much O2 as H2S for
complete reaction of both reactants. We don’t have that much O2, so O2 must be the
limiting reactant that will control the masses of each product produced.
d.
0.1563 mol O2 ×
2 mol SO 2 64.07 g SO 2
×
= 6.67 g SO2
3 mol O 2
1 mol SO 2
0.1563 mol O2 ×
2 mol H 2 O 18.02 g H 2 O
×
= 1.88 g H2O
3 mol O 2
1 mol H 2 O
3AgNO3(aq) + Al(s) → 3Ag(s) + Al(NO3)3(aq)
Molar masses: AgNO3, 169.9 g; Al, 26.98 g; Ag, 107.9 g; Al(NO3)3, 213.0 g
5.00 g AgNO3 ×
5.00 g Al ×
1 mol
= 0.02943 mol AgNO3
169.9 g
1 mol
= 0.1853 mol Al
26.98 g
According to the balanced chemical equation, we would need three moles of AgNO3 for
every mole of Al for complete reaction of both reactants. We in fact have fewer moles of
AgNO3 than aluminum, so AgNO3 must be the limiting reactant. The amount of product
produced is calculated from the number of moles of the limiting reactant present:
0.02943 mol AgNO3 ×
3 mol Ag
107.9 g Ag
×
= 3.18 g Ag
3 mol AgNO3 1 mol Ag
0.02943 mol AgNO3 ×
1 mol Al(NO3 )3 213.0 g Al(NO3 )3
×
= 2.09 g
3 mol AgNO3
1 mol Al(NO3 )3
181
Chapter 9:
47.
Chemical Quantities
Before any calculations are attempted, the equations must be balanced.
a.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
molar masses: C3H8, 44.09 g; O2, 32.00 g; CO2, 44.01 g; H2O, 18.02 g
10.0 g C3H8 ×
10.0 g O2 ×
1 mol
= 0.2268 mol C3H8
44.09 g
1 mol
= 0.3125 mol O2
32.00 g
For 0.2268 mol C3H8, the amount of O2 that would be needed is
0.2268 mol C3H8 ×
5 mol O 2
= 1.134 mol O2
1 mol C3 H8
Because we do not have this amount of O2, then O2 is the limiting reactant.
b.
0.3125 mol O2 ×
3 mol CO 2 44.01 g CO 2
×
= 8.25 g CO2
5 mol O 2
1 mol CO 2
0.3125 mol O2 ×
4 mol H 2 O 18.02 g H 2 O
×
= 4.51 g H2O
5 mol O 2
1 mol H 2 O
2Al(s) + 3Cl2(g) → 2AlCl3(s)
molar masses: Al, 26.98 g; Cl2, 70.90 g; AlCl3, 133.3 g
10.0 g Al ×
1 mol
= 0.3706 mol Al
26.98 g
10.0 g Cl2 ×
1 mol
= 0.1410 mol Cl2
70.90 g
For 0.1410 mol Cl2(g), the amount of Al(s) required is
0.1410 mol Cl2 ×
2 mol Al
= 0.09400 mol Al
3 mol Cl2
We have far more than this amount of Al(s) present, so Cl2(g) must be the limiting
reactant that will control the amount of AlCl3 which forms.
0.1410 mol Cl2 ×
c.
2 mol AlCl3 133.3 g AlCl3
×
= 12.5 g AlCl3
3 mol Cl2
1 mol AlCl3
2NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)
molar masses: NaOH, 40.00 g; CO2, 44.01 g; Na2CO3, 106.0 g; H2O, 18.02 g
10.0 g NaOH ×
10.0 g CO2 ×
1 mol
= 0.2500 mol NaOH
40.00 g
1 mol
= 0.2272 mol CO2
44.01 g
182
Chapter 9:
Chemical Quantities
Without having to calculate, according to the balanced chemical equation, we would need
twice as many moles of NaOH as CO2 for complete reaction. For the amounts calculated
above, there is not nearly enough NaOH present for the amount of Cl2 used: NaOH is the
limiting reactant.
d.
0.2500 mol NaOH ×
1 mol Na 2 CO3 106.0 g Na 2 CO3
×
= 13.3 g Na2CO3
2 mol NaOH
1 mol Na 2 CO3
0.2500 mol NaOH ×
1 mol H 2 O 18.02 g H 2 O
×
= 2.25 g H2O
2 mol NaOH 1 mol H 2 O
NaHCO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
molar masses: NaHCO3, 84.01 g; HCl, 36.46 g; NaCl, 58.44 g; H2O, 18.02 g; CO2, 44.01
g
10.0 g NaHCO3 ×
10.0 g HCl ×
1 mol
= 0.1190 mol NaHCO3
84.01 g
1 mol
= 0.2742 mol HCl
36.46 g
Because the coefficients of NaHCO3(s) and HCl(aq) are both one in the balanced
chemical equation for the reaction, there is not enough NaHCO3 present to react with the
amount of HCl present: the 0.1190 mol NaHCO3 present is the limiting reactant. Because
all the coefficients of the products are also each one, then if 0.1190 mol NaHCO3 reacts
completely (with 0.1190 mol HCl), 0.1190 mol of each product will form.
0.1190 mol NaCl ×
48.
a.
58.44 g
= 6.95 g NaCl
1 mol
0.1190 mol H2O ×
18.02 g
= 2.14 g H2O
1 mol
0.1190 mol CO2 ×
44.01 g
= 5.24 g CO2
1 mol
CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)
Molar masses: CS2, 76.15 g; O2, 32.00 g; CO2, 44.01 g
1.00 g CS2 ×
1.00 g O2 ×
1 mol
= 0.01313 mol CS2
76.15 g
1 mol
= 0.03125 mol O2
32.00 g
From the balanced chemical equation, we would need three times as much oxygen as
carbon disulfide for complete reaction of both reactants. We do not have this much
oxygen, and so oxygen must be the limiting reactant.
0.03125 mol O2 ×
1 mol CO 2 44.01 g CO 2
×
= 0.458 g CO2
3 mol O 2
1 mol CO 2
183
Chapter 9:
b.
Chemical Quantities
2NH3(g) + CO2(g) → CN2H4O(s) + H2O(l)
Molar masses: NH3, 17.03 g; CO2, 44.01 g; H2O, 18.02 g
1.00 g NH3 ×
1 mol
= 0.05872 mol NH3
17.03 g
1.00 g CO2 ×
1 mol
= 0.02272 mol CO2
44.01 g
The balanced chemical equation tells us that we would need twice as many moles of
ammonia as carbon dioxide for complete reaction of both reactants. We have more than
this amount of ammonia present, so the reaction will be limited by the amount of carbon
dioxide present.
0.02272 mol CO2 ×
c.
1 mol H 2 O 18.02 g H 2 O
×
= 0.409 g H2O
1 mol CO 2 1 mol H 2 O
H2(g) + MnO2(s) → MnO(s) + H2O(l)
Molar masses: H2, 2.016 g; MnO2, 86.94 g; H2O, 18.02 g
1 mol
= 0.496 mol H2
2.016 g
1.00 g H2 ×
1.00 g MnO2 ×
1 mol
= 0.0115 mol MnO2
86.94 g
Because the coefficients of both reactants in the balanced chemical equation are the same,
we would need equal amounts of both reactants for complete reaction. Therefore,
manganese(IV) oxide must be the limiting reactant and controls the amount of product
obtained.
0.0115 mol MnO2 ×
d.
1 mol H 2 O 18.02 g H 2 O
×
= 0.207 g H2O
1 mol MnO 2 1 mol H 2 O
I2(s) + Cl2(g) → 2ICl(g)
Molar masses: I2, 253.8 g; Cl2, 70.90 g; ICl, 162.35 g
1.00 g I2 ×
1 mol
= 0.00394 mol I2
253.8 g
1.00 g Cl2 ×
1 mol
= 0.0141 mol Cl2
70.90 g
From the balanced chemical equation, we would need equal amounts of I2 and Cl2 for
complete reaction of both reactants. As we have much less iodine than chlorine, iodine
must be the limiting reactant.
0.00394 mol I2 ×
2 mol ICl 162.35 g ICl
×
= 1.28 g ICl
1 mol I 2
1 mol ICl
184
Chapter 9:
49.
a.
Chemical Quantities
UO2(s) + 4HF(aq) → UF4(aq) + 2H2O(l)
UO2 is the limiting reactant; 1.16 g UF4, 0.133 g H2O
b.
NaNO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2HNO3(aq)
NaNO3 is the limiting reactant; 0.836 g Na2SO4; 0.741 g HNO3
c.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
HCl is the limiting reactant; 1.87 g ZnCl2; 0.0276 g H2
d.
B(OH)3(s) + 3CH3OH(l) → B(OCH3)3(s) + 3H2O(l)
CH3OH is the limiting reactant; 1.08 g B(OCH3)3; 0.562 g H2O
50.
a.
CO(g) + 2H2(g) → CH3OH(l)
CO is the limiting reactant; 11.4 mg CH3OH
b.
2Al(s) + 3I2(s) → 2AlI3(s)
I2 is the limiting reactant; 10.7 mg AlI3
c.
Ca(OH)2(aq) + 2HBr(aq) → CaBr2(aq) + 2H2O(l)
HBr is the limiting reactant; 12.4 mg CaBr2; 2.23 mg H2O
d.
2Cr(s) + 2H3PO4(aq) → 2CrPO4(s) + 3H2(g)
H3PO4 is the limiting reactant; 15.0 mg CrPO4; 0.309 mg H2
51.
Pb(C2H3O2)2(aq) + H2O(l) + CO2(g) → PbCO3(s) + 2HC2H3O2(aq)
molar masses: Pb(C2H3O2)2, 325.3 g; CO2, 44.01 g; PbCO3, 267.21 g
1.25 g Pb(C2H3O2)2 ×
5.95 g CO2 ×
1 mol
= 0.00384 mol Pb(C2H3O2)2
325.3 g
1 mol
= 0.135 mol CO2
44.01 g
Pb(C2H3O2)2 is the limiting reactant, which determines the yield of product.
0.00384 mol Pb(C2H3O2)2 ×
52.
1 mol PbCO3
267.21 g PbCO3
×
= 1.03 g PbCO3
1 mol Pb ( C2 H3O 2 )2
1 mol PbCO3
CuO(s) + H2SO4(aq) → CuSO4(aq) + H2O(l)
molar masses: CuO, 79.55 g; H2SO4, 98.09 g
2.49 g CuO ×
1 mol CuO
= 0.0313 mol CuO
79.55 g CuO
5.05 g H2SO4 ×
1 mol H 2SO 4
= 0.0515 mol H2SO4
98.09 g H 2SO 4
Since the reaction is of 1:1 stoichiometry, CuO must be the limiting reactant since it is present in
the lesser amount on a molar basis.
185
Chapter 9:
53.
Chemical Quantities
PbO(s) + C(s) → Pb(l) + CO(g)
molar masses: PbO, 223.2 g; C, 12.01 g; Pb, 207.2 g
50.0 × 103 g PbO ×
50.0 × 103 g C ×
1 mol
= 224.0 mol PbO
223.2 g
1 mol
= 4163 mol C
12.01 g
PbO is the limiting reactant.
224.0 mol PbO ×
54.
1 mol Pb 207.2 g Pb
×
= 4.64 × 104 g = 46.4 kg Pb
1 mol PbO 1 mol Pb
4Fe(s) + 3O2(g) → 2Fe2O3(s)
Molar masses: Fe, 55.85 g; Fe2O3, 159.7 g
1.25 g Fe ×
1 mol
= 0.0224 mol Fe present
55.85 g
Calculate how many mol of O2 are required to react with this amount of Fe
0.0224 mol Fe ×
3 mol O 2
= 0.0168 mol O2
4 mol Fe
Because we have more O2 than this, Fe must be the limiting reactant.
0.0224 mol Fe ×
55.
2 mol Fe2 O3 159.7 g Fe 2 O3
×
= 1.79 g Fe2O3
4 mol Fe
1 mol Fe 2 O3
Ag+(aq) + Cl–(aq) → AgCl(s)
molar masses: AgNO3, 169.91 g; NaCl, 58.44 g
The number of moles of silver ion present will be the same as the number of moles of silver
nitrate taken because each formula unit of silver nitrate contains one silver ion.
1.15 g AgNO3 ×
1 mol AgNO3
= 0.00677 mol AgNO3 = 0.00677 mol Ag+ ion
169.91 g AgNO3
The number of moles of chloride ion present will be the same as the number of moles of sodium
chloride taken, because each formula unit of sodium chloride contains one sodium ion.
5.45 g NaCl ×
1 mol NaCl
= 0.0933 mol NaCl = 0.0933 mol Cl– ion
58.44 g NaCl
Because the balanced chemical equation indicates a 1:1 stoichiometry for the reaction, there is not
nearly enough silver ion present (0.00677 mol) to precipitate the amount of chloride ion in the
sample (0.0933 mol).
56.
CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaCl(aq)
molar masses: CaCl2, 110.98 g; Na2SO4, 142.05 g
186
Chapter 9:
5.21 g CaCl2 ×
Chemical Quantities
1 mol CaCl2
= 0.0469 mol CaCl2 = 0.0469 mol Ca2+ ion
110.98 g CaCl2
4.95 g Na2SO4 ×
1 mol Na 2SO 4
= 0.0348 mol Na2SO4 = 0.0348 mol SO42– ion
142.05 g Na 2SO 4
Because the balanced chemical equation indicates a 1:1 stoichiometry for the reaction, there is not
nearly enough sulfate ion present (0.0348 mol) to precipitate the amount of calcium ion in the
sample (0.0469 mol). Sodium sulfate (sulfate ion) is the limiting reactant. Calcium chloride
(calcium ion) is present in excess.
57.
BaO2(s) + 2HCl(aq) → H2O2(aq) + BaCl2(aq)
molar masses: BaO2, 169.3 g; HCl, 36.46 g; H2O2, 34.02 g
1.50 g BaO2 ×
1 mol
= 8.860 × 10–3 mol BaO2
169.3 g
25.0 mL solution ×
0.680 g HCl ×
0.0272 g HCl
= 0.680 g HCl
1 mL solution
1 mol
= 1.865 × 10–2 mol HCl
36.46 g
BaO2 is the limiting reactant.
8.860 × 10–3 mol BaO2 ×
58.
1 mol H 2 O 2 34.02 g H 2 O 2
×
= 0.301 g H2O2
1 mol BaO2 1 mol H 2 O 2
SiO2(s) + 3C(s) → 2CO(g) + SiC(s)
molar masses: SiO2, 60.09 g; SiC, 40.10 g; 1.0 kg = 1.0 × 103 g
1.0 × 103 g SiO2 ×
1 mol
= 16.64 mol SiO2
60.09 g
From the balanced chemical equation, if 16.64 mol of SiO2 were to react completely (an excess of
carbon is present), then 16.64 mol of SiC should be produced (the coefficients of SiO2 and SiC
are the same).
16.64 mol SiC ×
40.01 g
= 6.7 × 102 g SiC = 0.67 kg SiC
1 mol
59.
The theoretical yield represents the yield we calculate from the stoichiometry of the reaction and
the masses of reactants taken for the experiment. The actual yield is what is actually obtained in
an experiment. The percent yield is the ratio of what is actually obtained to the theoretical amount
that could be obtained, converted to a percent basis.
60.
If the reaction is performed in a solvent, the product may have a substantial solubility in the
solvent; the reaction may come to equilibrium before the full yield of product is achieved (see
Chapter 16); loss of product may occur through operator error.
187
Chapter 9:
Chemical Quantities
actual yield
1.23 g
× 100 =
× 100 = 85.4%
theoretical yield
1.44 g
61.
Percent yield =
62.
2KClO3(s) → 2KCl(s) + 3O2(g)
molar mass: KClO3, 122.55 g; O2, 32.00 g
4.74 g KClO3 ×
1 mol KClO3
= 0.0387 mol KClO3
122.55 g KClO3
0.0387 mol KClO3 ×
32.00 g O 2
= 1.86 g O2 theoretical yield
1 mol O 2
0.0580 mol O2 ×
% yield =
63.
3 mol O 2
= 0.0580 mol O2
2 mol KClO3
1.51 g actual
× 100 = 81.3% of theory
1.86 g theoretical
S8(s) + 8Na2SO3(aq) + 40H2O(l) → 8Na2S2O3=5H2O
molar masses: S8, 256.6 g; Na2SO3, 126.1 g; Na2S2O3=5H2O, 248.2 g
3.25 g S8 ×
1 mol
= 0.01267 mol S8
256.6 g
1 mol
= 0.1039 mol Na2SO3
126.1 g
13.1 g Na2SO3 ×
S8 is the limiting reactant.
0.01267 mol S8 ×
8 mol Na 2S2 O3. 5H 2 O
= 0.1014 mol Na2S2O3⋅5H2O
1 mol S8
0.1014 mol Na2S2O3=5H2O ×
Percent yield =
64.
248.2 g Na 2S2 O3. 5H 2 O
= 25.2 g Na2S2O3⋅5H2O
1 mol Na 2S2 O3. 5H 2 O
actual yield
5.26 g
× 100 =
× 100 = 20.9%
theoretical yield
25.2 g
2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(g)
molar masses: LiOH, 23.95 g; CO2, 44.01 g
155 g LiOH ×
1 mol CO 2 44.01 g CO 2
1 mol LiOH
×
×
= 142 g CO2
23.95 g LiOH 2 mol LiOH 1 mol CO 2
As the cartridge has only absorbed 102 g CO2 out of a total capacity of 142 g CO2, the cartridge
has absorbed
102 g
× 100 = 71.8% of its capacity.
142 g
188
Chapter 9:
65.
Chemical Quantities
Xe(g) + 2F2(g) → XeF4(s)
molar masses: Xe, 131.3 g; F2, 38.00 g; XeF4, 207.3 g
130. g Xe ×
1 mol
= 0.9901 mol Xe
131.3 g
100. g F2 ×
1 mol
= 2.632 mol F2
38.00 g
Xe is the limiting reactant.
0.9901 mol Xe ×
Percent yield =
66.
1 mol XeF4 207.3 g XeF4
×
= 205 g XeF4
1 mol Xe
1 mol XeF4
actual yield
145 g
× 100 =
× 100 = 70.7 % of theory
theoretical yield
205 g
Ba2+(aq) + SO42–(aq) → BaSO4(s)
molar masses: SO42–, 96.07 g; BaCl2, 208.2 g; BaSO4, 233.4 g
1.12 g SO42– ×
1 mol
= 0.01166 mol SO42–
96.07 g
5.02 g BaCl2 ×
1 mol
= 0.02411 mol BaCl2 = 0.02411 mol Ba2+
208.2 g
SO42– is the limiting reactant.
0.01166 mol SO42– ×
Percent yield =
67.
1 mol BaSO 4 233.4 g BaSO 4
×
= 2.72 g BaSO4
1 mol SO 4 2 −
1 mol BaSO 4
actual yield
2.02 g
× 100 =
× 100 = 74.3%
theoretical yield
2.72 g
Ca(HCO3)2(aq) → CaCO3(s) + CO2(g) + H2O(l)
millimolar masses: Ca(HCO3)2, 162.1 mg; CaCO3, 100.1 mg
2.0 × 10–3 mg Ca(HCO3)2 ×
1 mmol
= 1.23 × 10–5 mmol Ca(HCO3)2
162.1 mg
1.23 × 10–5 mmol Ca(HCO3)2 ×
1.23 × 10–5 mmol ×
68.
1 mmol CaCO3
= 1.23 × 10–5 mmol CaCO3
1 mmol Ca(HCO3 ) 2
100.1 mg
= 1.2 × 10–3 mg = 1.2 × 10–6 g CaCO3
1 mmol
NaCl(aq) + NH3(aq) + H2O(l) + CO2(s) → NH4Cl(aq) + NaHCO3(s)
molar masses: NH3, 17.03 g; CO2, 44.01 g; NaHCO3, 84.01 g
189
Chapter 9:
Chemical Quantities
10.0 g NH3 ×
1 mol
= 0.5872 mol NH3
17.03 g
15.0 g CO2 ×
1 mol
= 0.3408 mol CO2
44.01 g
CO2 is the limiting reactant.
0.3408 mol CO2 ×
1 mol NaHCO3
= 0.3408 mol NaHCO3
1 mol CO 2
0.3408 mol NaHCO3 ×
69.
84.01 g
= 28.6 g NaHCO3
1 mol
Fe(s) + S(s) → FeS(s)
molar masses: Fe, 55.85 g; S, 32.07 g; FeS, 87.92 g
1 mol
= 0.0940 mol Fe
55.85 g
5.25 g Fe ×
12.7 g S ×
1 mol
= 0.396 mol S
32.07 g
Fe is the limiting reactant.
0.0940 mol Fe ×
70.
1 mol FeS 87.92 g FeS
×
= 8.26 g FeS produced
1 mol Fe
1 mol FeS
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)
molar masses: glucose, 180.2 g; CO2, 44.01 g
1.00 g glucose × = 5.549 × 10–3 mol glucose
5.549 × 10–3 mol glucose ×
3.33 × 10–2 mol CO2 ×
71.
6 mol CO 2
= 3.33 × 10–2 mol CO2
1 mol glucose
44.01 g
= 1.47 g CO2
1 mol
Cu(s) + S(s) → CuS(s)
molar masses: Cu, 63.55 g; S, 32.07 g; CuS, 95.62 g
31.8 g Cu ×
50.0 g S ×
1 mol
= 0.5004 mol Cu
63.55 g
1 mol
= 1.559 mol S
32.07 g
Cu is the limiting reactant.
0.5004 mol Cu ×
1 mol CuS
= 0.5004 mol CuS
1 mol Cu
190
Chapter 9:
0.5004 mol CuS ×
% yield =
72.
Chemical Quantities
95.62 g
= 47.8 g CuS
1 mol
40.0 g
× 100 = 83.7%
47.8 g
Ba2+(aq) + SO42–(aq) → BaSO4(s)
millimolar ionic masses: Ba2+, 137.3 mg; SO42–, 96.07 mg; BaCl2, 208.2 mg
150 mg SO42– ×
1 mmol
= 1.56 millimol SO42–
96.07 mg
As barium ion and sulfate ion react on a 1:1 stoichiometric basis, then 1.56 millimol of barium
ion is needed, which corresponds to 1.56 millimol of BaCl2
1.56 millimol BaCl2 ×
73.
208.2 mg
= 325 milligrams BaCl2 needed
1 mmol
mass of Cl– present = 1.054 g sample ×
10.3 g Cl−
= 0.1086 g Cl–
100.0 g sample
molar masses: Cl–, 35.45 g; AgNO3, 169.9 g; AgCl, 143.4 g
0.1086 g Cl– ×
1 mol
= 3.063 × 10–3 mol Cl–
35.45 g
3.063 × 10–3 mol Cl– ×
1 mol AgNO3
= 3.063 × 10–3 mol AgNO3
−
1 mol Cl
3.063 × 10–3 mol AgNO3 ×
3.063 × 10–3 mol Cl– ×
1 mol AgCl
= 3.063 × 10–3 mol AgCl
1 mol Cl−
3.063 × 10–3 mol AgCl ×
74.
a.
169.9 g
= 0.520 g AgNO3 required
1 mol
143.4 g
= 0.439 g AgCl produced
1 mol
UO2(s) + 4HF(aq) → UF4(aq) + 2H2O(l)
One molecule (formula unit) of uranium(IV) oxide will combine with four molecules of
hydrofluoric acid, producing one uranium(IV) fluoride molecule and two water
molecules. One mole of uranium(IV) oxide will combine with four moles of hydrofluoric
acid to produce one mole of uranium(IV) fluoride and two moles of water.
b.
2NaC2H3O2(aq) + H2SO4(aq) → Na2SO4(aq) + 2HC2H3O2(aq)
Two molecules (formula units) of sodium acetate react exactly with one molecule of
sulfuric acid, producing one molecule (formula unit) of sodium sulfate and two molecules
of acetic acid. Two moles of sodium acetate will combine with one mole of sulfuric acid,
producing one mole of sodium sulfate and two moles of acetic acid.
191
Chapter 9:
c.
Chemical Quantities
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
One magnesium atom will react with two hydrochloric acid molecules (formula units) to
produce one molecule (formula unit) of magnesium chloride and one molecule of
hydrogen gas. One mole of magnesium will combine with two moles of hydrochloric
acid, producing one mole of magnesium chloride and one mole of gaseous hydrogen.
d.
B2O3(s) + 3H2O(l) → 2B(OH)3(aq)
One molecule of diboron trioxide will react exactly with three molecules of water,
producing two molecules of boron trihydroxide (boric acid). One mole of diboron
trioxide will combine with three moles of water to produce two moles of boron
trihydroxide (boric acid).
75.
False. For 0.40 mol of Mg(OH)2 to react, 0.80 mol of HCl will be needed. According to the
balanced equation, for a given amount of Mg(OH)2, twice as many moles of HCl is needed.
76.
⎛ 5 mol O 2 ⎞
For O2: ⎜
⎟
⎝ 1 mol C3 H8 ⎠
77.
a.
b.
c.
d.
⎛ 3 mol CO 2 ⎞
For CO2: ⎜
⎟
⎝ 1 mol C3 H8 ⎠
⎛ 4 mol H 2 O ⎞
For H2O: ⎜
⎟
⎝ 1 mol C3 H8 ⎠
2H2O2(l) → 2H2O(l) + O2(g)
0.50 mol H2O2 ×
2 mol H 2 O
= 0.50 mol H2O
2 mol H 2 O 2
0.50 mol H2O2 ×
1 mol O 2
= 0.25 mol O2
2 mol H 2 O 2
2KClO3(s) → 2KCl(s) + 3O2(g)
0.50 mol KClO3 ×
2 mol KCl
= 0.50 mol KCl
2 mol KClO3
0.50 mol KClO3 ×
3 mol O 2
= 0.75 mol O2
2 mol KClO3
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
0.50 mol Al ×
2 mol AlCl3
= 0.50 mol AlCl3
2 mol Al
0.50 mol Al ×
3 mol H 2
= 0.75 mol H2
2 mol Al
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
0.50 mol C3H8 ×
3 mol CO 2
= 1.5 mol CO2
1 mol C3 H8
0.50 mol C3H8 ×
4 mol H 2 O
= 2.0 mol H2O
1 mol C3 H8
192
Chapter 9:
78.
a.
NH3(g) + HCl(g) → NH4Cl(s)
molar mass of NH3 = 17.01 g
1.00 g NH3 ×
1 mol
= 0.0588 mol NH3
17.01 g
0.0588 mol NH3 ×
b.
1 mol NH 4 Cl
= 0.0588 mol NH4Cl
1 mol NH 3
CaO(s) + CO2(g) → CaCO3(s)
molar mass CaO = 56.08 g
1.00 g CaO ×
1 mol
= 0.0178 mol CaO
56.08 g
0.0178 mol CaO ×
c.
1 mol CaCO3
= 0.0178 mol CaCO3
1 mol CaO
4Na(s) + O2(g) → 2Na2O(s)
molar mass Na = 22.99 g
1.00 g Na ×
1 mol
= 0.0435 mol Na
22.99 g
0.0435 mol Na ×
d.
2 mol Na 2 O
= 0.0217 mol Na2O
4 mol Na
2P(s) + 3Cl2(g) → 2PCl3(l)
molar mass P = 30.97 g
1.00 g P ×
1 mol
= 0.0323 mol P
30.97 g
0.0323 mol P ×
79.
a.
molar mass CuSO4 = 159.6 g
4.21 g CuSO4 ×
b.
2 mol PCl3
= 0.0323 mol PCl3
2 mol P
1 mol
= 0.0264 mol CuSO4
159.6 g
molar mass Ba(NO3)2 = 261.3 g
7.94 g Ba(NO3)2 ×
c.
1 mol
= 0.0304 mol Ba(NO3)2
261.3 g
molar mass water = 18.02 g; 1.24 mg = 0.00124 g
0.00124 g ×
1 mol
= 6.88 × 10–5 mol H2O
18.02 g
193
Chemical Quantities
Chapter 9:
d.
Chemical Quantities
molar mass W = 183.9 g
9.79 g W ×
e.
molar mass S = 32.07 g; 1.45 lb = 1.45(454) = 658 g
658 g S ×
f.
1 mol
= 5.32 × 10–2 mol W
183.9 g
1 mol
= 20.5 mol S
32.07 g
molar mass C2H5OH = 46.07 g
4.65 g C2H5OH ×
g.
molar mass C = 12.01 g
12.01 g C ×
80.
a.
1 mol
= 1.00 mol C
12.01 g
molar mass HNO3 = 63.0 g
5.0 mol HNO3 ×
b.
1 mol
= 0.101 mol C2H5OH
46.07 g
63.0 g
= 3.2 × 102 g HNO3
1 mol
molar mass Hg = 200.6 g
0.000305 mol Hg ×
c.
200.6 g
= 0.0612 g Hg
1 mol
molar mass K2CrO4 = 194.2 g
2.31 × 10–5 mol K2CrO4 ×
d.
molar mass AlCl3 = 133.3 g
10.5 mol AlCl3 ×
e.
133.3 g
= 1.40 × 103 g AlCl3
1 mol
molar mass SF6 = 146.1 g
4.9 × 104 mol SF6 ×
f.
146.1 g
= 7.2 × 106 g SF6
1 mol
molar mass NH3 = 17.01 g
125 mol NH3 ×
g.
194.2 g
= 4.49 × 10–3 g K2CrO4
1 mol
17.01 g
= 2.13 × 103 g NH3
1 mol
molar mass Na2O2 = 77.98 g
0.01205 mol Na2O2 ×
77.98 g
= 0.9397 g Na2O2
1 mol
194
Chapter 9:
81.
Before any calculations are done, the equations must be balanced.
a.
BaCl2(aq) + H2SO4(aq) → BaSO4(s) + 2HCl(aq)
0.145 mol BaCl2 ×
b.
1 mol H 2SO 4
= 0.145 mol H2SO4
1 mol BaCl2
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
0.145 mol AgNO3 ×
c.
1 mol NaCl
= 0.145 mol NaCl
1 mol AgNO3
Pb(NO3)2(aq) + Na2CO3(aq) → PbCO3(s) + 2NaNO3(aq)
0.145 mol Pb(NO3)2 ×
d.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
0.145 mol C3H8 ×
82.
1 mol Na 2 CO3
= 0.145 mol Na2CO3
1 mol Pb(NO3 ) 2
5 mol O 2
= 0.725 mol O2
1 mol C3 H8
2SO2(g) + O2(g) → 2SO3(g)
molar masses: SO2, 64.07 g; SO3, 80.07 g; 150 kg = 1.5 × 105 g
1.5 × 105 g SO2 ×
83.
1 mol
= 2.34 × 103 mol SO2
64.07 g
2.34 × 103 mol SO2 ×
2 mol SO3
= 2.34 × 103 mol SO3
2 mol SO 2
2.34 × 103 mol SO3 ×
80.07 g
= 1.9 × 105 g SO3 = 1.9 × 102 kg SO3
1 mol
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
molar masses: ZnS, 97.45 g; SO2, 64.07 g; 1.0 × 102 kg = 1.0 × 105 g
1.0 × 105 g ZnS ×
84.
1 mol
= 1.026 × 103 mol ZnS
97.45 g
1.026 × 103 mol ZnS ×
2 mol SO 2
= 1.026 × 103 mol SO2
2 mol ZnS
1.026 × 103 mol SO2 ×
64.07 g
= 6.6 × 104 g SO2 = 66 kg SO2
1 mol
2Na2O2(s) + 2H2O(l) → 4NaOH(aq) + O2(g)
molar masses: Na2O2, 77.98 g; O2, 32.00 g
3.25 g Na2O2 ×
1 mol
= 0.0417 mol Na2O2
77.98 g
195
Chemical Quantities
Chapter 9:
Chemical Quantities
0.0417 mol Na2O2 ×
0.0209 mol O2 ×
85.
1 mol O 2
= 0.0209 mol O2
2 mol Na 2 O 2
32.00 g
= 0.667 g O2
1 mol
Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)
millimolar masses: Cu, 63.55 mg; AgNO3, 169.9 mg
1.95 mg AgNO3 ×
1 mmol
= 0.01148 mmol AgNO3
169.9 mg
0.01148 mmol AgNO3 ×
0.005740 mmol Cu ×
86.
1 mmol Cu
= 0.005740 mmol Cu
2 mmol AgNO3
63.55 g
= 0.365 mg Cu
1 mol
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
molar masses: Zn, 65.38 g; H2, 2.016 g
2.50 g Zn ×
87.
1 mol
= 0.03824 mol Zn
65.38 g
0.03824 mol Zn ×
1 mol H 2
= 0.03824 mol H2
1 mol Zn
0.03824 mol H2 ×
2.016 g
= 0.0771 g H2
1 mol
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
molar masses: C2H2, 26.04 g; O2, 32.00 g; 150 g = 1.5 × 102 g
1.5 × 102 g C2H2 ×
1 mol
= 5.760 mol C2H2
26.04 g
5.760 mol C2H2 ×
5 mol O 2
= 14.40 mol O2
2 mol C2 H 2
14.40 mol O2 ×
88.
a.
32.00 g
= 4.6 × 102 g O2
1 mol
2Na(s) + Br2(l) → 2NaBr(s)
molar masses: Na, 22.99 g; Br2, 159.8 g; NaBr, 102.9 g
5.0 g Na ×
1 mol
= 0.2175 mol Na
22.99 g
196
Chapter 9:
5.0 g Br2 ×
Chemical Quantities
1 mol
= 0.03129 mol Br2
159.8 g
Intuitively, we would suspect that Br2 is the limiting reactant, because there is much less
Br2 than Na on a mole basis. To prove that Br2 is the limiting reactant, the following
calculation is needed:
0.03129 mol Br2 ×
2 mol Na
= 0.06258 mol Na.
1 mol Br2
Clearly, there is more Na than this present, so Br2 limits the reaction extent and the
amount of NaBr formed.
0.03129 mol Br2 ×
2 mol NaBr
= 0.06258 mol NaBr
1 mol Br2
0.06258 mol NaBr ×
b.
102.9 g
= 6.4 g NaBr
1 mol
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
molar masses: Zn, 65.38 g; Cu, 63.55 g; ZnSO4, 161.5 g; CuSO4, 159.6 g
5.0 g Zn ×
1 mol
= 0.07648 mol Zn
65.38 g
5.0 g CuSO4 ×
1 mol
= 0.03132 mol CuSO4
159.6 g
As the coefficients of Zn and CuSO4 are the same in the balanced chemical equation, an
equal number of moles of Zn and CuSO4 would be needed for complete reaction. There is
less CuSO4 present, so CuSO4 must be the limiting reactant.
0.03132 mol CuSO4 ×
1 mol ZnSO 4
= 0.03132 mol ZnSO4
1 mol CuSO 4
0.03132 mol ZnSO4 ×
161.5 g
= 5.1 g ZnSO4
1 mol
0.03132 mol CuSO4 ×
1 mol Cu
= 0.03132 mol Cu
1 mol CuSO 4
0.03132 mol Cu ×
c.
63.55 g
= 2.0 g Cu
1 mol
NH4Cl(aq) + NaOH(aq) → NH3(g) + H2O(l) + NaCl(aq)
molar masses: NH4Cl, 53.49 g; NaOH, 40.00 g; NH3, 17.03 g; H2O, 18.02 g;
NaCl, 58.44 g
5.0 g NH4Cl ×
1 mol
= 0.09348 mol NH4Cl
53.49 g
197
Chapter 9:
Chemical Quantities
5.0 g NaOH ×
1 mol
= 0.1250 mol NaOH
40.00 g
As the coefficients of NH4Cl and NaOH are both one in the balanced chemical equation
for the reaction, an equal number of moles of NH4Cl and NaOH would be needed for
complete reaction. There is less NH4Cl present, so NH4Cl must be the limiting reactant.
As the coefficients of the products in the balanced chemical equation are also all one, if
0.09348 mol of NH4Cl (the limiting reactant) reacts completely, then 0.09348 mol of
each product will be formed.
0.09348 mol NH3 ×
17.03 g
= 1.6 g NH3
1 mol
0.09348 mol H2O ×
18.02 g
= 1.7 g H2O
1 mol
0.09348 mol NaCl ×
d.
58.44 g
= 5.5 g NaCl
1 mol
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
molar masses: Fe2O3, 159.7 g; CO, 28.01 g; Fe, 55.85 g; CO2, 44.01 g
5.0 g Fe2O3 ×
5.0 g CO ×
1 mol
= 0.03131 mol Fe2O3
159.7 g
1 mol
= 0.1785 mol CO
28.01 g
Because there is considerably less Fe2O3 than CO on a mole basis, let’s see if Fe2O3 is the
limiting reactant.
0.03131 mol Fe2O3 ×
3 mol CO
= 0.09393 mol CO
1 mol Fe2 O3
As there is 0.1785 mol of CO present, but we have determined that only 0.09393 mol CO
would be needed to react with all the Fe2O3 present, then Fe2O3 must be the limiting
reactant. CO is present in excess.
89.
a.
0.03131 mol Fe2O3 ×
2 mol Fe
55.85 g Fe
×
= 3.5 g Fe
1 mol Fe 2 O3 1 mol Fe
0.03131 mol Fe2O3 ×
3 mol CO 2 44.01 g CO 2
×
= 4.1 g CO2
1 mol Fe2 O3 1 mol CO 2
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
molar masses: C2H5OH, 46.07 g; O2, 32.00 g; CO2, 44.01 g
25.0 g C2H5OH ×
25.0 g O2 ×
1 mol
= 0.5427 mol C2H5OH
46.07 g
1 mol
= 0.7813 mol O2
32.00 g
198
Chapter 9:
Chemical Quantities
As there is less C2H5OH present on a mole basis, see if this substance is the limiting
reactant.
0.5427 mol C2H5OH ×
3 mol O 2
= 1.6281 mol O2.
1 mol C2 H 5OH
From the above calculation, C2H5OH must not be the limiting reactant (even though there
is a smaller number of moles of C2H5OH present) because more oxygen than is present
would be required to react completely with the C2H5OH present. Oxygen is the limiting
reactant.
0.7813 mol O2 ×
b.
2 mol CO 2 44.01 g CO 2
×
= 22.9 g CO2
3 mol O 2
1 mol CO 2
N2(g) + O2(g) → 2NO(g)
molar masses: N2, 28.02 g; O2, 32.00 g; NO, 30.01 g
25.0 g N2 ×
1 mol
= 0.8922 mol N2
28.02 g
25.0 g O2 ×
1 mol
= 0.7813 mol O2
32.00 g
As the coefficients of N2 and O2 are the same in the balanced chemical equation for the
reaction, an equal number of moles of each substance would be necessary for complete
reaction. There is less O2 present on a mole basis, so O2 must be the limiting reactant.
0.7813 mol O2 ×
c.
2 mol NO 30.01 g NO
×
= 46.9 g NO
1 mol O2
1 mol NO
2NaClO2(aq) + Cl2(g) → 2ClO2(g) + 2NaCl(aq)
molar masses: NaClO2, 90.44 g; Cl2, 70.90 g; NaCl, 58.44 g
25.0 g NaClO2 ×
25.0 g Cl2 ×
1 mol
= 0.2764 mol NaClO2
90.44 g
1 mol
= 0.3526 mol Cl2
70.90 g
See if NaClO2 is the limiting reactant.
0.2764 mol NaClO2 ×
1 mol Cl2
= 0.1382 mol Cl2
2 mol NaClO 2
As 0.2764 mol of NaClO2 would require only 0.1382 mol Cl2 to react completely (and
since we have more than this amount of Cl2), then NaClO2 must indeed be the limiting
reactant.
0.2764 mol NaClO2 ×
2 mol NaCl
58.44 g NaCl
×
= 16.2 g NaCl
2 mol NaClO 2 1 mol NaCl
199
Chapter 9:
d.
Chemical Quantities
3H2(g) + N2(g) → 2NH3(g)
molar masses: H2, 2.016 g; N2, 28.02 g; NH3, 17.03 g
25.0 g H2 ×
1 mol
= 12.40 mol H2
2.016 g
25.0 g N2 ×
1 mol
= 0.8922 mol N2
28.02 g
See if N2 is the limiting reactant.
0.8922 mol N2 ×
3mol H 2
= 2.677 mol H2
1 mol N 2
N2 is clearly the limiting reactant, because there is 12.40 mol H2 present (a large excess).
0.8922 mol N2 ×
90.
2 mol NH 3 17.03 g NH 3
×
= 30.4 g NH3
1 mol N 2
1 mol NH 3
N2H4(l) + O2(g) → N2(g) + 2H2O(g)
molar masses: N2H4, 32.05 g; O2, 32.00 g; N2, 28.02 g; H2O, 18.02 g
20.0 g N2H4 ×
20.0 g O2 ×
1 mol
= 0.624 mol N2H4
32.05 g
1 mol
= 0.625 mol O2
32.00 g
The two reactants are present in nearly the required ratio for complete reaction (due to the 1:1
stoichiometry of the reaction and the very similar molar masses of the substances). We will
consider N2H4 as the limiting reactant in the following calculations.
91.
0.624 mol N2H4 ×
1 mol N 2
28.02 g N 2
×
= 17.5 g N2
1 mol N 2 H 4 1 mol N 2
0.624 mol N2H4 ×
2 mol H 2 O 18.02 g H 2 O
×
= 22.5 g H2O
1 mol N 2 H 4 1 mol H 2 O
Total quantity of H2S = 50. L ×
1.5 × 10-5 g
= 7.5 × 10–4 g H2S
1L
8H2S(aq) + 8Cl2(aq) → 16HCl(aq) + S8(s)
molar masses: H2S, 34.09 g; Cl2, 70.90 g; S8, 256.6 g
7.5 × 10–4 g H2S ×
1.0 g Cl2 ×
1 mol
= 2.20 × 10–5 mol H2S
34.09 g
1 mol
= 1.41 × 10–2 mol Cl2
70.90 g
There is a large excess of chlorine present compared to the amount of Cl2 that would be needed to
react with all the H2S present in the water sample: H2S is the limiting reactant for the process.
200
Chapter 9:
2.20 × 10–5 mol H2S ×
92.
12.5 g theory ×
1 mol S8
256.6 g S8
×
= 7.1 × 10–4 g S8 removed
8 mol H 2S 1 mol S8
40 g actual
= 5.0 g
100 g theory
201
Chemical Quantities